Originally posted by emu:

There are 2 ways to measure power.

The first is to take a voltage reading with the impedance known. Let's say you have a 2 ohm load and you measured it to be 4.1 ohms at 50 Hz. Measure the max voltage at 50 Hz, now power = voltage * voltage / load(4.1)

The second and better way is to measure voltage and current simultaneously. With a clamp-on ampmeter, burp the amp and write down the voltage and current you got.
power = voltage * current
load = voltage / current

As for rms power, forget about it. Most meters are rms to start with. But the scientific way to measure rms power is take peak power and divide that by 2.
Originally posted by ShadowStar:

Most meters that you use will measure RMS AC voltage, not peak AC voltage. The quickest way to tell is to stick the probes in a light socket. If you measure 115-120v, then your meter measures RMS and you're in good shape, and can ignore everything else. If they measure something higher than 130, then your meter measures peak and you should multiply the voltage and current and then cut that number in half if finding out how much power you are getting matters.
Originally posted by ShadowStar:

RMS AC power is the amount of AC that is equivalent to some amount of DC.

Lets say you have a 10vDC battery. This battery will induce a current in a 1 ohm resistive load of 10 amps. That is 100 watts.

Now, look at the top half of a sine wave. (AC voltage.) If you use a 10vAC signal, which swings from 0 volts to 10 volts at the very tippity top of the wave, then the current will also go from 0 amps to 10 amps at the very tippity top of the wave. This means that the power level will go from 0 watts to 100 watts at the very top of the wave. The area under the power level is actually how much power is delivered.

So, we use the RMS measure of AC voltage, which is .707 times the voltage, which gives us the amount of voltage which will induce .707 times the current.

(.707*V)*(.707*I) = .5*P

Where P is the power.

RMS ac power gives you how much heating power you're applying to the coil.